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195=-5t^2+10t+210
We move all terms to the left:
195-(-5t^2+10t+210)=0
We get rid of parentheses
5t^2-10t-210+195=0
We add all the numbers together, and all the variables
5t^2-10t-15=0
a = 5; b = -10; c = -15;
Δ = b2-4ac
Δ = -102-4·5·(-15)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{400}=20$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-20}{2*5}=\frac{-10}{10} =-1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+20}{2*5}=\frac{30}{10} =3 $
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